If the shortest distance between the lines (x- )/(2)=(y-4)/(3)=(z-3)/(4) and (x-2)/(4)=(y-…

Question

If the shortest distance between the lines $\frac{x-\lambda }{2}=\frac{y-4}{3}=\frac{z-3}{4}$ and $\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-7}{8}$ is $\frac{13}{\sqrt{29}}$, then a value of $\lambda $ is :

Accepted Answer

$$\begin{aligned}
& \vec{l}_1=2 \hat{i}+3 \hat{j}+4 \hat{k} \
& \vec{l}_2=4 \hat{i}+6 \hat{j}+8 \hat{k}
\end{aligned}$$

$$S . D .=\frac{|(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times((\lambda-2) \hat{i}-4 \hat{k})|}{|2 \hat{i}+3 \hat{j}+4 \hat{k}|}$$

$$\begin{aligned}
& \left|\[\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 4 \ \lambda-2 & 0 & -4 \end{array}\end{aligned}$$\right|=-12 \hat{i}-\hat{j}(-8-4 \lambda+8)+\hat{k}(6-3 \lambda) \
\& =-12 \hat{i}+4 \lambda \hat{j}+(6-3 \lambda) \hat{k} \
\& \frac{\sqrt{144+16 \lambda\^2+(6-3 \lambda)\^2}}{\sqrt{29}}=\frac{13}{\sqrt{29}} \
\& 144+16 \lambda\^2+(6-3 \lambda)\^2=169 \
\& \Rightarrow 16 \lambda\^2+9 \lambda\^2+36-36 \lambda+144-169=0 \
\]

$$\begin{aligned}
& \Rightarrow 25 \lambda^2-36 \lambda+11=0 \
& \Rightarrow 25 \lambda^2-25 \lambda-11 \lambda+11=0 \
& \Rightarrow(25 \lambda-11)(\lambda-1)=0 \
& \Rightarrow \lambda=1, \frac{11}{25}
\end{aligned}$$

Solutions · 1